# [SOLVED] Binomial distribution problem

• Mar 31st 2010, 12:13 AM
downthesun01
[SOLVED] Binomial distribution problem
According to Chemical Engineering Progress, approximately 30% of all pipework failures in chemical plants are caused by operator error.

Suppose, for a particular plant, that out of a random sample of 20 pipework failures, exactly 5 are operational errors. Do you feel that the 30% figure stated above applies to this plant? Comment.

I did the math and

\$\displaystyle p(X=5)= 0.1789\$

The book says that this answer means that the 30% figure is acceptable, but I don't understand why. I would have thought that the 30% figure would be acceptable if \$\displaystyle p(X=5)\$ was close to 1. Can someone help me understand this? Thanks
• Mar 31st 2010, 11:28 AM
Anonymous1
Probability is a rather elusive subject. Off the bat you have \$\displaystyle 5/20 = .25\$ operational failures so this looks good. the probability of this EXACT result is around \$\displaystyle .18. \$ Think of it this way, we could have \$\displaystyle 1 \$ operational error, \$\displaystyle 2,...,20.\$ etc.. and the sum of ALL these possibilities is \$\displaystyle 1.\$ So when you are saying what is the probability of EXACTLY \$\displaystyle 5\$ operational errors. A probability of \$\displaystyle \approx .18\$ is definitely one of the highest possible outcomes.

The key is to understand that there is \$\displaystyle 20\$ possible outcomes. If each of the outcomes were eqaully likely, with probability \$\displaystyle .18.\$ Then the sum of all possibilities would be around \$\displaystyle 4\$ and not \$\displaystyle 1.\$ So the outliers like \$\displaystyle 1\$ operational error out of \$\displaystyle 20\$ will have a very small probibility, and the out come with the highest probability would be \$\displaystyle 6\$ operational errors. Why is this?

Do you get what I'm saying?
• Mar 31st 2010, 08:42 PM
downthesun01
I see what you mean about p(X=5) = being the second highest probability and p(X=6) being the highest. I guess my question now is what exactly constitutes the idea that the 30% figure is acceptable? I mean it's the SECOND highest probability, not the highest. To me, because it's not THE highest probability, I would say that the 30% isn't acceptable. It just seems kind of relative.
• Mar 31st 2010, 08:58 PM
Anonymous1
Quote:

Originally Posted by downthesun01
I see what you mean about p(X=5) = being the second highest probability and p(X=6) being the highest. I guess my question now is what exactly constitutes the idea that the 30% figure is acceptable? I mean it's the SECOND highest probability, not the highest. To me, because it's not THE highest probability, I would say that the 30% isn't acceptable. It just seems kind of relative.

25% operational failures is very close to 30% for such a small sample size. Random process, are just that: random. If you flip a coin 5 times and get 5 heads in a row, you may question the fairness of the coin. The fact is rare events happen, just at a lower frequency. Now, if you flipped the coin 100^100 times you could expect to see heads 50% of the time.

At this plant we have 5 operational errors when our expectation was 6. Standard deviation takes the relative component out of the equation. I would bet that 5/20 errors is less than one standard deviation below the mean: a very acceptable number. Have you got 75% (average) on every single test you've taken? Or do you deviate, earning +-1 standard dev and the occasional 2.
• Mar 31st 2010, 11:12 PM
mr fantastic
Quote:

Originally Posted by downthesun01
According to Chemical Engineering Progress, approximately 30% of all pipework failures in chemical plants are caused by operator error.

Suppose, for a particular plant, that out of a random sample of 20 pipework failures, exactly 5 are operational errors. Do you feel that the 30% figure stated above applies to this plant? Comment.

I did the math and

\$\displaystyle p(X=5)= 0.1789\$

The book says that this answer means that the 30% figure is acceptable, but I don't understand why. I would have thought that the 30% figure would be acceptable if \$\displaystyle p(X=5)\$ was close to 1. Can someone help me understand this? Thanks

To answer the question you first have to decide on a value such that if the probability is below that value you will reject the claim. A value of 1 is completely unreasonable.